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strike through incorrect assertion.

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edited Sep 8 2010 at 4:57

Different take: For the application you want, however, you don't need such a general theory. In your application, the order of every element divides the order of the distinguished element. The disadvantage is that this cannot be expressed as an identity, so you do not have a variety of algebras. You do not have the nice constructions such as coproducts, products, free objects, etc. a priori. On the other hand, I think that the classification is pretty straightforward: every such group can be expressed as a product of cyclic groups, $C_{n_1}\times\cdots\times C_{n_k}$, with $n_1|n_2|\cdots|n_k$, and with distinguished element a specific generator of the largest cyclic factor in the case where $n_k>0$, ~~and an arbitrary nonzero element of the last cyclic factor if $n_k=0$~~ and a torsion element plus a nonzero element of the last cyclic factor if $n_k=0$. However, this decomposition is not decomposing the objects into something which is a direct product of pointed abelian groups (the direct summands other than the largest ones do not contain the distinguished element). The proof in the finite case can be done the usual way; for an abelian $p$-group $A$, if $x$ is an element of maximal order then $A$ is isomorphic to $A/\langle x\rangle \oplus \langle x\rangle$, so you can start by picking the distinguished element; otherwise, consider the $p$-parts and think of them as quotients of the original pointed group, with distinguished element the image of the distinguished element; then put them together in the usual way, and the distinguished element will correspond to a generator of the largest cyclic factor. In the infinite case, it gets a bit more complicated, but it should go through: take $A/A^{tor}$, with distinguished element the image of the distinguished element of $A$; this is a direct sum of cyclic groups of infinite order as an abelian group, and the subgroup generated by the distinguished element is a submodule of rank $1$. So if I remember my modules over PIDs correctly, there is a basis for $A/A^{tor}$ of the form $x_1,\ldots,x_r$, in which $1$ is a multiple of $x_r$; then deal with the torsion part as you would with a normal finite abelian group. Edit: No, the infinite case does not go through like this; see comment. You can get it so that it will be of the form $(a,(0,\ldots,0,d))$, where $a$ is in the torsion part and $(0,\ldots,0,d)$ is an element of the torsion free part with $d>0$, but you may not be able to "get rid" of the $a$. This is problematical.

fix silly mistake on decomposing an abelian group

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edited Sep 8 2010 at 0:36

Arturo Magidin
2,691●1●12●25

Different take: For the application you want, however, you don't need such a general theory. In your application, the order of every element divides the order of the distinguished element. The disadvantage is that this cannot be expressed as an identity, so you do not have a variety of algebras. You do not have the nice constructions such as coproducts, products, free objects, etc. a priori. On the other hand, I think that the classification is pretty straightforward: every such group can be expressed as a product of cyclic groups, $C_{n_1}\times\cdots\times C_{n_k}$, with $n_1|n_2|\cdots|n_k$, and with distinguished element a specific generator of the largest cyclic factor . in the case where $n_k>0$, and an arbitrary nonzero element of the last cyclic factor if $n_k=0$. However, this decomposition is not decomposing the objects into something which is a direct product of pointed abelian groups (the direct summands other than the largest ones do not contain the distinguished element). The proof in the finite case can be done the usual way, since if you have a finitely generated ; for an abelian group, and $p$-group $A$, if $x$ is an element of maximal order of $A$, then $A$ is isomorphic to $A/\langle x\rangle \oplus \langle x\rangle$, so you can start by picking the distinguished element; otherwise, consider the $p$-parts and think of them as quotients of the original pointed group, with distinguished element the image of the distinguished element; then put them together in the usual way, and the distinguished element will correspond to a generator of the largest cyclic factor. So In the infinite case, it gets a bit more complicated, but it should go through: take $A/A^{tor}$, with distinguished element the image of the distinguished element of $A$; this is a direct sum of cyclic groups of infinite order as an abelian group, and the subgroup generated by the distinguished element first and go from is a submodule of rank $1$. So if I remember my modules over PIDs correctly, there is a basis for $A/A^{tor}$ of the form $x_1,\ldots,x_r$, in which $1$ is a multiple of $x_r$; then deal with the torsion part as you would with a normal finite abelian group.

fix the 1-generator pointed case.

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edited Sep 8 2010 at 0:21

Arturo Magidin
2,691●1●12●25

This category includes the entire theory of abelian groups as the class of those groups in which $0=1$; so any classification theory will be much harder than the one for abelian groups. For example, the $1$ generator pointed abelian groups: every $1$-generator pointed abelian group is really a $2$-generator abelian group: it is generated by the given generator plus the distinguished element $1$. So suppose $x$ is the given generator. If $\langle 1\rangle\subseteq\langle x\rangle$, then you get one distinct an isomorphism class for each pair type of the form $(C,C')$, where $C$ is a cyclic group and $C'$ is a subgroup of $C$ (corresponding $C$ corresponds to the subgroup cyclic group generated by $x$, and $C'$ to the distinguished element; abelian subgroup generated by $1$). But these do not exhaust all possibilities. You also get groups correspond in which $\langle x\rangle\cap\langle 1\rangle = \{0\}$; if the orders are coprime, you are back to the previous case$C'=0$). , but if the orders are not coprime then the description is a bit more complicated. And then the case where they intersect nontrivially but neither contains the other, and so on. As you can see, things get complicated even in this simplest of cases...

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answered Sep 7 2010 at 23:53

Arturo Magidin
2,691●1●12●25

I have some general comments.

Summary: such a general theory would be much harder than that for abelian groups (and in fact, would contain the classification of finitely generated abelian groups as a special case), and you would lose a lot of the "good" properties of the category of abelian groups along the way. Many constructions (such as the coproduct) would be different from what we are used to. The advantage is that you can fit into the general context of Universal Algebra, so all of the familiar "abstract nonsense" theory will apply (suitably interpreted). On the other hand, if you restrict to the case you are actually interested in, in which the order of each element divides the order of the distinguished element, then you end up with a much simpler theory, but it falls outside the usual confines of Universal Algebra. You can write down an easy description, but it's not going to be well-behaved relative to homomorphisms.

Some details. To consider the general theory from the point of view of Universal Algebra, you are looking at algebras of type (2,1,0,0), where the binary operation is denoted $+$ (with the usual infix notation), the unary operation is "$-$", one of the zero-ary operations is denoted "$0$", and the second unary operation is denoted "$1$". The operations are required to satisfy the obvious identities: $(a+b)+c = a+(b+c)$, $a+b=b+a$, $a+0=a$, $a+(-a)=0$, and no identity is required of the second unary operation $1$. Homomorphisms are required to respect addition, inversion, map $0$ to $0$, and map $1$ to $1$. The class of abelian groups (as usual) is a variety of algebras of this type, characterized by the identity $1=0$. So you have an embedding from abelian groups to "pointed abelian groups", and this embedding has left adjoint which corresponds to taking the quotient modulo the subgroup generated by $1$. You also have a "forgetful" functor from "pointed abelian groups" to abelian groups, obtained by forgetting about the pointed element;

This class, however, is no longer an abelian category. I'm pretty sure of the following, but I could have made some mistakes: The product of two pointed abelian groups $(A,1)$ and $(B,1')$ is, as Henry suggested in the comments, the abelian group $A\times B$ with distinguished element $(1,1')$, and the structure projections are the canonical projections. The free pointed abelian group on a set $X$ will be the direct sum of $|X|+1$ copies of $\mathbb{Z}$, with the generator of the extra copy being the distinguished element. The coproduct of $(A,1)$ and $(B,1')$ will be the quotient of $A\oplus B\oplus\mathbb{Z}$ modulo the subgroup generated by $(1,0,1)$ and $(0,1',1)$, and the distinguished element is the image of $(0,0,1)$. The canonical immersions go are the canonical maps into the direct sum followed by the quotient map.

This category includes the entire theory of abelian groups as the class of those groups in which $0=1$; so any classification theory will be much harder than the one for abelian groups. For example, the $1$ generator pointed abelian groups: you get one distinct isomorphism class for each pair $(C,C')$, where $C$ is a cyclic group and $C'$ is a subgroup of $C$ (corresponding to the subgroup generated by the distinguished element; abelian groups correspond to the case $C'=0$). Two pointed abelian groups that are isomorphic as pointed abelian groups will be isomorphic as abelian groups, but you can have to pointed abelian groups whose forgetful image is isomorphic, but that are not isomorphic as pointed abelian groups; for example, $(\mathbb{Z},1)$ and $(\mathbb{Z},2)$ are not isomorphic as pointed abelian groups.

The advantage is that you have the entire weaponry from Universal Algebra at your disposal, as well as two reasonably well-behaved functors to abelian groups: the forgetful functor, and the adjoint of the inclusion of the variety of abelian groups to the variety of pointed abelian groups. The disadvantage is that it is unfamiliar, and more complicated than the usual theory of abelian groups. (Think about how things get more difficult going form abelian groups to not-necessarily-abelian groups).

Different take: For the application you want, however, you don't need such a general theory. In your application, the order of every element divides the order of the distinguished element. The disadvantage is that this cannot be expressed as an identity, so you do not have a variety of algebras. You do not have the nice constructions such as coproducts, products, free objects, etc. a priori. On the other hand, I think that the classification is pretty straightforward: every such group can be expressed as a product of cyclic groups, $C_{n_1}\times\cdots\times C_{n_k}$, with $n_1|n_2|\cdots|n_k$, and with distinguished element a specific generator of the largest cyclic factor. However, this decomposition is not decomposing the objects into something which is a direct product of pointed abelian groups (the direct summands other than the largest ones do not contain the distinguished element). The proof can be done the usual way, since if you have a finitely generated abelian group, and $x$ is an element of maximal order of $A$, then $A$ is isomorphic to $A/\langle x\rangle \oplus \langle x\rangle$. So take the subgroup generated by the distinguished element first and go from there. The advantage is that there's the description of all such groups. There are some obvious properties, such as: you can only map from $(A,1)$ to $(B,1)$ if the order of the distinguished element of $A$ is a multiple of the order of the distinguished element of $B$; the map must send the distinguished generator of the largest cyclic factor of the expression for $A$ to the distinguished generator of the largest cyclic factor of $B$. Among the disadvantages is that it is not a "decomposition" in the sense of expressing the pointed group in terms of "smaller" pointed groups, and so it may not lead to any sensible "simplification", just a "description", which is probably not what you want. A more sensible simplification would require you to identify all "product irreducible" pointed abelian groups, and this might be pretty difficult.